It is not hard to see that, given one sample from a univariate unit-variance Gaussian $X\sim \mathcal{N}(\mu,1)$ with unknown $\mu$ s.t. $0<|\mu|\leq 1$, one can simulate one draw from a "Bernoulli" random variable $Y\in\{-1,1\}$ with $|\mathbb{E}[Y]| = \Theta(|\mu|)$. (Basically, just take $Y=\operatorname{sign}(X)$.)

Is it possible to do the converse (even allowing extra randomness, independent of the sample)? That is:

Given one realization of some random variable $Y\in\{-1,1\}$ with (unknown) $\mathbb{E}[Y] = \mu$, output $X\sim \mathcal{N}(\nu,1)$ s.t. $|\nu| = \Theta(|\mu|)$.

I conjecture that to be impossible (I have some mild contrived evidence), but I don't really see how one would go about proving that.


1 Answer 1


Suppose you had such a randomized procedure that takes a value in $\{-1,1\}$ and outputs a real number. Let $P$ and $Q$ be the output distribution on input $+1$ and $-1$ respectively.

Consider the extreme case of $\mu = +1$. In this case $Y = +1$ for sure, and you are outputting a sample from $P$, which means that $P$ should be an $\mathcal{N}(\nu, 1)$ distribution for some $\nu$ with $|\nu|$ being $\Omega(1)$. Similarly, the other extreme of $\mu=-1$ tells us that $Q$ is a $\mathcal{N}(\nu', 1)$ distribution for some $\nu'$ with $|\nu'|$ being $\Omega(1)$.

But now, for $\mu=0$, you are outputting the uniform mixture of two normals, which cannot be a normal distribution unless $\nu=\nu'$. In the latter case, you get $\mathcal{N}(\nu, 1)$ for some $\nu$ bounded away from zero.

Thus you cannot satisfy this requirement simultaneously for $\mu\in \{-1, 0, 1\}$.

  • 1
    $\begingroup$ That was fast—thanks! $\endgroup$
    – Clement C.
    Dec 16, 2020 at 0:52

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